Codeforces Round #285 (Div. 2) (A、B、C、D）

標籤：

A：就根據題意計算比較一下即可

B：從每個起點往後走一遍走到底，輸出即可，字串直接map映射掉

C：類似拓撲排序，從臨接個數為1的入隊，那麼臨接Xor和，其實就是他的上一個結點，因為他只臨接了一個結點，這樣利用拓撲排序，當一個結點的度數為1的時候入隊即可，注意要判斷一下度數0的情況，直接continue

D：利用樹狀數組去求這種大的全排列數，其實一個全排列 ，可以看成a1 * (n - 1)! + a2 * (n - 2)!....，那麼其實只要處理出每一項的係數，然後在由係數就可以求出變換後的全排列了，那麼求係數這一步，只需要把兩個序列的係數加起來，然後進行進位操作，最後在轉化回去即可，這個過程要利用樹狀數組來維護，因為每個位置都要查詢，當前剩下數字中，比該數字小的數字個數

代碼：

A：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;double a, b, c, d;int main() {    scanf("%lf%lf%lf%lf", &a, &b, &c, &d);    double a1 = max(3 * a / 10, a - a / 250 * c);    double a2 = max(3 * b / 10, b - b / 250 * d);    if (a1 < a2) printf("Vasya\n");    else if (a1 > a2) printf("Misha\n");    else printf("Tie\n");    return 0;}

B：

#include <cstdio>#include <cstring>#include <string>#include <map>using namespace std;int n, hn, vis[2005], to[2005];map<string, int> hash;char a[25], b[25];char out[2005][25];int get(char *str) {    if (hash.count(str))return hash[str];    strcpy(out[hn], str);    hash[str] = hn++;    return hash[str];}int dfs(int u) {    while (1) {if (to[u] == -1) break;u = to[u];    }    return u;}int main() {    scanf("%d", &n);    hn = 0;    memset(to, -1, sizeof(to));    while (n--) {scanf("%s%s", a, b);int u = get(a), v = get(b);to[u] = v;vis[v] = 1;    }    int tot = 0;    for (int i = 0; i < hn; i++)if (!vis[i]) {    tot++;}    printf("%d\n", tot);    for (int i = 0; i < hn ;i++)if (!vis[i]) {    printf("%s %s\n", out[i], out[dfs(i)]);}    return 0;}

C：

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <vector>using namespace std;typedef pair<int, int> pii;const int N = (1<<16) + 5;int n, du[N], s[N];vector<pii> ans;int main() {    scanf("%d", &n);    queue<int> Q;    for (int i = 0; i < n; i++) {scanf("%d%d", &du[i], &s[i]);if (du[i] == 1) Q.push(i);    }    while (!Q.empty()) {int u = Q.front();Q.pop();if (du[u] != 1) continue;ans.push_back(make_pair(u, s[u]));s[s[u]] ^= u;du[s[u]]--;if (du[s[u]] == 1) Q.push(s[u]);    }    int tot = ans.size();    printf("%d\n", tot);    for (int i = 0; i < tot; i++)printf("%d %d\n", ans[i].first, ans[i].second);    return 0;}

D：

#include <cstdio>#include <cstring>#define lowbit(x) (x&(-x))const int N = 200005;int n, a[N];int bit[N];void add(int x, int v) {    while (x <= n) {bit[x] += v;x += lowbit(x);    }}int query(int x) {    int ans = 0;    while (x) {ans += bit[x];x -= lowbit(x);    }    return ans;}int find(int x) {    int l = 1, r = n;    while (l < r) {int mid = (l + r) / 2;int tmp = query(mid);if (tmp < x) l = mid + 1;else r = mid;    }    return l;}int main() {    scanf("%d", &n);    memset(bit, 0, sizeof(bit));    for (int i = 1; i <= n; i++) add(i, 1);    int x;    for (int i = 1; i <= n; i++) {scanf("%d", &x); x++;int tmp = query(x) - 1;add(x, -1);a[i] += tmp;    }    for (int i = 1; i <= n; i++) add(i, 1);    for (int i = 1; i <= n; i++) {scanf("%d", &x); x++;int tmp = query(x) - 1;add(x, -1);a[i] += tmp;    }    for (int i = n; i >= 1; i--) {a[i - 1] += a[i] / (n - i + 1);a[i] %= (n - i + 1);    }    for (int i = 1; i <= n; i++) add(i, 1);    for (int i = 1; i <= n; i++) {int tmp = find(a[i] + 1);printf("%d ", tmp - 1);add(tmp, -1);    }    printf("\n");    return 0;}

Codeforces Round #285 (Div. 2) (A、B、C、D）